Kinetic Equation For Gases | KMT of Gases

The kinetic molecular theory of gases tells us about the kinetic equation for gases. This theory also explains the kinetic energy of gases and on what factor the kinetic energy of ideal gas depends. The kinetic molecular theory of gases is completely valid for ideal gases but, partially valid for non-ideal gases or real gases. Here are some scientists and their contributions are given below:

• Bernoulli (1738): Founder of the kinetic molecular theory of gases.
• Clausius (1857): Derive kinetic equation and deduced all the gas laws from KMT.
• Maxwell: Presented the law of distribution of velocities.
• Van der Waal: Corrected the pressure and volume factor in the ideal gas equation to make it applicable to real gases.

Before we derive the kinetic equation, it’s very important to understand the postulates of the kinetic molecular theory of gases.

Postulates of Kinetic Molecular Theory

1. Gas is made up of small particles called atoms or molecules. Gases like helium, neon, and argon have mono-atomic molecules.
2. Molecules of small gas are identical in all respects (having the same mass, shape, and size).
3. Molecules are constantly in random motion along a straight line. (assumption)

Let a molecule of gas is moving with speed u. The components of u are in three directions u_x, u_y, and u_z.

It assumes that the probability of movement of a gas molecule in the X-direction, Y-direction, and Z-direction is the same so,

u_x = u_y = u_z

The magnitude of u:

u = u_x i + u_y j + u_z k

u = √u_x² + u_y² + u_z²

u² = u_x² + u_y² + u_z²

If; u_x = u_y = u_z

u² = 3u_x²

4. All the collisions of gas molecules among themselves and with the wall of the container are elastic in nature. Elastic in nature means there is no loss in momentum (momentum remains conserved) and kinetic energy (kinetic energy remains the same before and after collision).

5. The pressure of a gas is due to the collision of molecules with the wall of the container.

We have a container containing a gas molecule, a molecule of gas is moving with momentum mv: 

When a gas molecule collides with the wall of the container, its momentum is changed (-mv). The momentum of a gas molecule is the same but opposite in direction. The gas molecule exerts a force on the wall of the container and the change in momentum is due to force so, the force exerted by a gas molecule per unit area is called pressure. The more the number of molecules collides, the more will be the pressure.

6. The kinetic energy of gas (ideal) depends only and only upon absolute temperature. The kinetic energy of gas does not depend upon the nature of the gas. It also does not depend upon the pressure.

The nature of gas means if we take one mole of hydrogen gas (ideal) and 1 mole of Oxygen gas (ideal) at 27 degrees Celsius, the kinetic energy of the gas is the same.

7. The volume occupied by gas molecules is negligible when compared to the volume of gas. (assumption)

8. There is no intermolecular force of attraction among molecules. (assumption)

9. There is no effect of gravity on the motion of the molecules of a gas.

Kinetic Equation for Gases

On the basis of kinetic theory, it is possible to derive a relation between the pressure and volume of gas in the form of an equation.

PV = 1/3 mn u_rms²

Where,

• P = Pressure
• V = Volume
• m = mass of one molecule of gas
• n = number of molecules
• u_rms² = Root mean square velocity of molecules

What is u_rms²?

Let us have an nth number of gas particles having different velocities u₁, u₂, u₃,…, u_n.

u = u₁ + u₂ + u₃ + …… + u_n

Squaring both sides

u² = u₁² + u₂² + u₃² + …… + u_n²

Taking average (mean)

u² = u₁² + u₂² + u₃² + …… + u_n² / N

where, N = total number of molecules

Taking under root both sides

√u² = √u₁² + u₂² + u₃² + …… + u_n² / N

u_rms = √u₁² + u₂² + u₃² + …… + u_n² / N

This is called root mean square velocity.

Derivation of Kinetic Equation For Gases

Let we have a gas container having n number of gas molecules of direction x, y, and z, and length L.

Consider a small molecule of a gas with mass m and velocity u₁. As we know that the velocity of a molecule has three components but we assume only u_x component and we are also considering only one side of the container which is also along the x-direction.

How to pressure exerted by a small molecule of gas when colliding with the wall of the container?

• The collision of a gas molecule is elastic so kinetic energy remains the same. (Before and after collision)
• The momentum is changed.

Change in Momentum in Each Collision

Intial momentum = P_i = mu_x1

Final momentum = P_f = -mu_x1

Change in momentum = Initial momentum – Final momentum

△P = mu_x1 – (- mu_x1)

△P = 2mu_x1 (△P in each collision)

The change in momentum is due to force

Time Taken For Each Collision

As the length of the side is “L” and the speed of the molecule is u_x1.

• Before collision: a molecule of a gas is moved toward a wall of the container.
• After collision: the molecule of a gas is moved away from the wall of the container.

So, the total length is 2L.

time = distance / speed

time = 2L / u_x1 (time taken for each collision)

Force Exerted on the Wall

Force exerted on wall = △P / △t = change in momentum / change in time

putting the values of △P and △t

Force exerted on wall = 2mu_x1 / 2L / u_x1

Force exerted on wall = 2mu_x1 × u_x1 / 2L

Force exerted on wall = mu_x1² / L ——- (1)

(Force applied on the wall by one molecule due to collision)

If we want to write in terms of u, then

u = u_x i + u_y j + u_z k

u_x = u_y = u_z

u² = u_x² i + u_y² j + u_z² k

u² = 3u_x²

u₁² = 3u_x1²

u_x1² = 1/3 u₁²

Put in eq (1)

Force exerted on wall = m/L × 1/3 u₁²

Force exerted on wall = m/3L × u₁² (Force due to one molecule)

Force due to two molecules

F₂= m/3L × u₂²

Force due to three molecules

F₃= m/3L × u₃²

Force due to nth molecules

F_n= m/3L × u_n²

F_net = F₁ + F₂ + F₃ + …… + F_n

F_net = m/3L × u₁² + m/3L × u₂² + m/3L × u₃² + ….. + m/3L × u_n²

F_net = m/3L ( u₁² + u₂² + u₃² + ….. + u_n²)

Multiply and divided bt “n”

F_net = m/3L ( u₁² + u₂² + u₃² + ….. + u_n² / n) × n

As, u_rms = √u₁² + u₂² + u₃² + …… + u_n² / N

u_rms²= u₁² + u₂² + u₃² + …… + u_n² / N

F_net = mn/3L × u_rms²

As,

Pressure = Force / Area

Pressure = mn/3L × u_rms² / A

Pressure = mn u_rms² / 3L A

As, A = L × W = L × L = L²

P = mn u_rms² / 3L × L²

P = mn u_rms² / 3L³

As, L³ = volume of cylinder = L × W × h = L × L × L = L³

P = mn u_rms² / 3 V

PV = 1/3 mn u_rms²

This is called Kinetic equation for gases.

PV = 1/3 mn u_rms² ——– (1)

As we know that the kinetic energy is equal to 1/2 mv²

K.E of one molecule = 1/2 mu₁²

K.E of two molecules = 1/2 mu₂²

K.E of three molecules = 1/2 mu₃²

K.E of nth molecules = 1/2 mu_n²

K.E of n molecules = 1/2 mu₁² + 1/2 mu₂² + 1/2 mu₃² + …… + 1/2 mu_n²

K.E of n molecules = 1/2 m (u₁² + u₂² + u₃² + …… + u_n² )

K.E of n molecules = 1/2 m (u₁² + u₂² + u₃² + …… + u_n² / n) × n

K.E of n molecules = 1/2 mn (u_rms²) ——– (2)

Divided (2) by (1)

K.E / PV = 1/2 mn (u_rms²) / 1/3 mn u_rms²

K.E / PV = 3/2

K.E = 3/2 PV ——– (3)

As,

PV = nRT

For one mole of an ideal gas n=1

PV = RT

Put of the value of PV in (3)

K.E = 3/2 RT (KE of one mole of an ideal gas)

The most important thing is that the kinetic energy of the gas is independent of its nature of the gas. It only depends on the temperature.

As we know,

K.E of one mole of an ideal gas = 3/2 RT

For n moles of an ideal gas

K.E of n moles of an ideal gas = 3/2 nRT

As 1 mole = N_A molecules = 6.02×10²³ molecules

K.E of one molecule = 3RT / 2N_A

Where R/N_A is Bolt’zmen constant (K)

K.E of one molecule = 3/2 KT

Note:

Basically, these derivatives are only true for monoatomic ideal gas. ( Helium, Neon, and Argon are monoatomic gases)

Deduction of Gas Laws From the Kinetic Equation of Gas

1. Boyle’s Law

According to the kinetic equation

PV = 1/3 mnu_rms² or

PV = (2/3) × 1/2mnu_rms² ——– (1)

Where,

• 1/2 mnu_rms² is the average kinetic energy of molecules.

According to the kinetic molecular theory of gases, the average kinetic energy of the molecule is directly proportional to the absolute temperature.

K.E ∝ T

1/2 mnu_rms² ∝ T

1/2 mnu_rms² = KT ——– (2)

From equation (1) and (2)

PV = 2/3 KT

At constant temperature, PV = K which is Boyle’s Law

2. Charles’s Law

As

PV = 2/3 KT

V = 2/3 KT/P

At constant pressure, 2/3 K/P is constant

Hence, V ∝ T which is Charles’s Law

3. Avogadro’s Law

As the kinetic equation of gas,

PV = 1/3 mnu_rms²

For gas (1); P₁V₁ = 1/3 m₁n₁u_1rms²

For gas (2); P₂V₂ = 1/3 m₂n₂u₂rms²

If the pressure and volume of two gases are equal, then

1/3 m₁n₁u_1rms² = 1/3 m₂n₂u₂rms² ———- (3)

If two gases are at the same temperature, the average kinetic energy per molecule will be the same.

1/2 m₁u₁² = 1/2 m₂u₂² ——– (4)

Divided equation (3) by (4)

n₁ = n₂ which is Avodagro’s Law.

4. Graham’s Law of Diffusion

According to the kinetic gas equation,

PV = 1/3 mnu_rms²

For one mole of gas n=N_A, we can write

PV = 1/3 mN_Au_rms²

Where N_A is Avogadro’s number (mN_A = M)

PV = 1/3 Mu_rms²

u_rms² = 3PV/M = 3P/M/V = 3P/d (d=M/V)

or u_rms² ∝ 1/d

u_rms ∝ √1/d

Now, “r” the rate of diffusion of the gas is directly proportional to the rms velocity of the molecules, thus

r ∝ √1/d which is Graham’s law of diffusion.

5. Dalton’s Law of Partial Pressure

If n₁ molecules of a gas, each of mass m₁ and velocity u₁, occupy a volume V, then pressure P exerted by these molecules is given by,

P₁ = 1/3 m₁n₁u_1rms/V

Similarly, pressure P₂ exerted by n₂ molecules of another gas, each of mass m₂ and velocity u₂ is given by,

P₂ = 1/3 m₂n₂u_2rms/V

In the same way, pressure P₃ of a third gas is given by

P₃ = 1/3 m₃n₃u_3rms/V

If all the three gases are kept together in another flask of the same volume and at the same temperature then total pressure P is given by,

P = P₁ = 1/3 m₁n₁u_1rms/V + 1/3 m₂n₂u_2rms/V + 1/3 m₃n₃u_3rms/V

P = P₁ + P₂ + P₃ which is Dalton’s Law of partial pressure.

Numericals of Kinetic Equation

Find the kinetic equation of 8g of methane at 27℃.

Temp = 27℃ = 27 + 273K = 300K

K.E_{(1 mole)} = 3/2 RT

For n moles,

K.E_{(n moles)} = 3/2 nRT

n = m/M = 8/16 = 0.5

K.E_{(n moles)} = 3/2 (0.5) × (8.314) × 300K

K.E = 1870.65 J/mol

Find the change in kinetic energy of one mole of an ideal gas when temperature changes by 50℃.

K.E = 3/2 RT

Intial K.E = K.E_i = 3/2 RT

Final K.E = K.E_f = 3/2 R (T+50℃)

△K.E = Final – Initial

△K.E = 3/2 R (T+50℃) – 3/2 RT

△K.E = 3/2 R (T+50 – T)

△K.E = 3/2 R × 50

△K.E = 3/2 × 8.314 ×50

△K.E = 610.05 J

At what temperature, the kinetic energy will be half of its value at 127℃?

Temp = 127℃ = 400K

K.E = 3/2 RT

K.E_(127℃) = 3/2 R × 400K

At what temperature “T” the kinetic energy will half

K.E_(T) = K.E_(127℃) / 2

3/2 RT = 3/2 R × 400K / 2

T = 400/2

T = 200K

Another method:

K.E ∝ T

When the temperature is half, the kinetic energy will behalf

Given temp = 127℃ = 400K

K.E = 400K

Temperature becomes half 200K

K.E/2 = 200k

We have a container containing 1 mole of helium with a molar mass of 4g. One size length (L) of the container is 2cm and they make 500 collisions with the wall each second. Find the temperature of the gas.

As helium gas makes 500 collisions in one second so,

Time for one second = 1/500 s ——– (1)

Time is taken for one collision = Distance / speed

Time is taken for one collision = 2L / u_rms ——– (2)

Compare (1) and (2)

1/500 = 2L / u_rms

1/500 = 2 × 2×10^-2 / u_ums

u_rms = (2 × 2×10^-2) × (500)

u_rms = 20 m/s

K.E of one molecule = 1/2 mu_rms² ——– (3)

K.E of one molecule = 3/2 KT ——– (4) where K=Bolt’zmen constant

Compare (3) and (4)

1/2 mu_rms² = 3/2 KT

mu_rms² = 3 KT

T = mu_rms² / 3K ——– (5)

m is the number of molecules

Molar mass of one mole of He = 4g

mass of one molecule of He = 4 amu

As we know that 1 amu = 1.67×10^-27kg

m = 4 × 1.67×10^-27

K = 1.38×10^-23 J/K (K is Bolt’zmen constant)

Putting all the values in (5)

T =4 × 1.67×10^-27 × (20)² / 3(1.38×10^-23)

T = 2.672×10^-24 / 4.414×10^-23

T = 0.0515 K